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Let $A$ be a matrix of order $n$ over a field $F$. Let the characteristic polynomial of the matrix $A$ be an irreducible polynomial in $F$, and let $M_{n}(F)$ be the set of $n$ order complex matrices over $F$.

(1):prove the matrix $A$ is invertible

This aswer I can prove it.

(2)Let $\sigma_{A}$ be a linear transformation in $M_{n}(F)$, such that $$\sigma_{A}(X)=A^{-1}X-XA^{-1},\forall X \in M_{n}(F)$$ Find $$\ker\sigma_{A}\cap \operatorname{Im}\sigma_{A}$$

My try: I can only prove this if the matrix $A$ is invertible. Thank you for you help.

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  • $\begingroup$ From the text it seems that $A$ must be invertible, or you can't define $\sigma_A$. Please let me know if my edit didn't change your question. $\endgroup$
    – egreg
    Nov 6, 2013 at 15:52
  • $\begingroup$ The transformation $\sigma_A $ is defined only when $ A $ is invertible. How can you use the fact that the characteristic polynomial of $ A $ is irreducible for that purpose? $\endgroup$
    – DKal
    Nov 6, 2013 at 16:00
  • $\begingroup$ Hello,this problem have two question:(1):prove $A$ is inverible. $\endgroup$
    – math110
    Nov 6, 2013 at 16:08
  • $\begingroup$ Why do you write with bold? egreg removed the blods in the first formulation and then you added them back... $\endgroup$ Nov 15, 2013 at 14:36

2 Answers 2

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Characteristic polynomial is irreducible implies that the eigenvalues of $A$ are distinct (and not contained in $F$). Changing $A$ by a conjugate we may assume that $A$ is diagonal over an extension of $F$. In that case the kernel of the map $\sigma_{A}$ is the set of all commuting matrices and they are only the diagonal ones. The image of $\sigma_{A}$ have trace zero. Thus the required intersection is any diagonal matrix with trace zero.

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(1) A polynomial without independent term is certainly reducible. Since the characteristic polynomial is irreducible, it must have a non-zero independent term. This means that $\det A$ is not zero and so $A$ is invertible.

(2) Since the characteristic polynomial is irreducible, it must coincide with the minimal polynomial. In this case, every matrix that commutes with $A$ must be a polynomial in $A$ (see for instance this question).

If $X\in \ker \sigma_A$ then $X$ commutes with $A$ and so is a polynomial in $A$.

I don't know right now what to make of $X\in \operatorname{im} \sigma_A$.

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